Question: Let $y=\cos(5x^2-x)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\sin(5x^2-x)$ (Choice B) B $(1-10x)\sin(5x^2-x)$ (Choice C) C $\sin(1-10x)\cos(5x^2-x)$ (Choice D) D $\cos(1-10x)$
$\cos(5x^2-x)$ is a trigonometric expression, but its argument isn't simply $x$. Therefore, it defines a composite trigonometric function. In other words, suppose $u(x)=5x^2-x$, then $y=\cos\Bigl(u(x)\Bigr)$. $\dfrac{dy}{dx}$ can be found using the following identity: $\dfrac{d}{dx}\left[\cos\Bigl(u(x)\Bigr)\right]=-{\sin\Bigl(u(x)\Bigr)}\cdot u'(x)$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\cos(5x^2-x) \\\\ &=\dfrac{d}{dx}\cos\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=5x^2-x} \\\\ &=-{\sin\Bigl(u(x)\Bigr)}\cdot u'(x) \\\\ &=-{\sin\Bigl(5x^2-x\Bigr)}\cdot (10x-1)&&\gray{\text{Substitute }u(x)\text{ back}} \\\\ &=(1-10x)\sin(5x^2-x) \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=(1-10x)\sin(5x^2-x)$.